Nonnoetherian Geometry

Home , Krull dimension

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Dimer algebras with the cancellation property are Calabi-Yau algebras, and their centers are 3-dimensional Gorenstein singularities. Non-cancellative dimer algebras, on the other hand, are not Calabi-Yau, and their centers are nonnoetherian. In contrast to their cancellative counterparts, very little is known about these algebras, despite the fact that almost all dimer algebras are non-cancellative. In [B2], we use the tools developed here to show that their centers are also 3-dimensional singularities, but with the strange property that they contain positive dimensional ‘smeared-out’ points. Furthermore, in [B4] we describe how this nonlocal geometry is reflected in the homology of their noncommutative residue fields. (2) Quantum entanglement and nonlocality. In quantum physics, two entangled particles can effect each instantaneously no matter how great their spatial separa- tion. In [B5, B6], we introduce a toy model where such entangled states are supported on nonnoetherian singularities of spacetime. Furthermore, we show that the noncommutative blowup of the singular support of an entangled Bell state leads to the notion that the nonlocal commutative spacetime that we observe emerges from an underlying local noncommutative spacetime [B5, Theorem C]. (3) Nonnoetherian geometry. In algebraic geometry, one may ask what is the geometry of a nonnoetherian algebra R? The usual answer is the affine scheme Spec R whose global sections are R. However, schemes are abstract objects, and we would like an answer to this question that we can visualize. We aim to give such an answer here. We now outline our main results. In Section 2 we introduce the idea of a nonlocal algebraic variety. To illustrate this notion, consider the algebra S = k[x, y] and its nonnoetherian subalgebra (1) R = k x, xy, xy2,..., = k + xS. The maximal ideal spectrum Max R of R may be viewed as 2-dimensional affine space 2 Ak = Max S with the line (x)= x =0 A2 Z { } ⊂ k identified as a single closed point. From this perspective, (x) is a 1-dimensional ‘smeared-out’ point of R, and therefore Max R is nonlocal. Z More generally, let S be an integral domain and a finitely generated k-algebra, and let R be a (possibly nonnoetherian) subalgebra of S. In order to capture the locus where Max R ‘looks like’ the variety Max S, we introduce the open subset U := n Max S R = S . S/R { ∈ | n∩R n} Theorem A. (Theorem 2.5.) Suppose US/R = . Then Max S and Max R are isomorphic on open dense subsets, and thus birationally6 ∅ equivalent. Furthermore, the Krull dimensions of R and S are equal.

In example (1), US/R is the complement to the subvariety (x). We generalize this example by showing that if R is generated by a subalgebraZ of S and an ideal I S, ⊂ NONNOETHERIAN GEOMETRY 3 then US/R contains the complement to the subvariety (I) in Max S. In addition, if Z c I is a non-maximal radical ideal of S and R = k + I, then US/R = (I) (Proposition 2.8). Z To formalize these notions, we introduce the following definitions. Definition B. (Definition 3.1.) A finitely generated k-algebra S is a depiction of a subalgebra R S if • ⊆ (i) the morphism ιS : Spec S Spec R, q q R, is surjective, (ii) for each n Max S, R →is noetherian7→ iff R∩ = S , and ∈ n∩R n∩R n (iii) US/R = . The geometric6 ∅ codimension or geometric height of p Spec R is the infimum • ∈ ght(p) := inf ht(q) q ι−1(p), S a depiction of R . | ∈ S The geometric dimension of p is the difference gdim p := dim R ght(p). − In example (1), R is depicted by S, and the geometric dimension of the closed point (x) of R is 1 (Example 3.11). The following theorem describes the fundamental Zgeometry of nonnoetherian algebras with finite Krull dimension. Theorem C. (Theorems 3.8, 3.12, 3.13, and Proposition 3.16.) Suppose R admits a depiction S and let p Spec R. Then ∈ ght(p) ht(p), ≤ with equality if there is some q Spec S for which q R = p and (q) US/R = . Furthermore, the following are∈ equivalent: ∩ Z ∩ 6 ∅ (1) R is noetherian. (2) US/R = Max S. (3) R = S. In particular, if R is noetherian, then its only depiction is itself. Finally, if Max R contains a closed point of positive geometric dimension, then R is nonnoetherian. Conversely, if R is an isolated nonnoetherian singularity, then Max R contains a closed point of positive geometric dimension. Consequently, if I is a radical ideal of a finitely generated k-algebra S, then the ring R = k + I will be nonnoetherian if and only if dim (I) 1 (Corollary 3.14). We conclude the section by showing that depictions whichZ are≥ minimal with respect to inclusion do not exist in general, and maximal depictions are not unique in general (Proposition 3.19). In Section 4, we study nonlocality in the context of noncommutative algebraic geometry. Let A be a finitely generated noncommutative k-algebra with center Z. We consider algebras with the following particularly nice matrix ring structure. 4 CHARLIE BEIL

Definition 1.1. An impression of A is a finitely generated commutative k-algebra B and an algebra monomorphism τ : A ֒ Md(B) such that (i) for each b in some open dense subset of Max B, the composition→ with the evaluation map ǫ A τ M (B) b M (B/b) −→ d −→ d is surjective, and (ii) the morphism Max B Max τ(Z), b b1d τ(Z), is surjective [B, Definition and Lemma 2.1]. → 7→ ∩ An impression is useful in part because it determines the center Z of A [B, Lemma 2.1]. Furthermore, if A is a finitely generated Z-module, then τ determines all simple A-module isoclasses of maximal k-dimension [B, Proposition 2.5]. Denote by Eji Md(k) the matrix with a 1 in the (ji)-th slot and zeros elsewhere. Let A = kQ/I be∈ a quiver algebra with vertex set Q = 1,...,d , and suppose A 0 { } admits an impression τ : A ֒ Md(B) such that τ(ei) = Eii for each i Q0. For p e Ae , denote byτ ¯(p) B→the single non-zero entry of the matrix τ(p∈); whence ∈ j i ∈ τ(p)=¯τ(p)Eji. For each i, j Q ,τ ¯ defines a k-linear mapτ ¯ : e Ae B. Set ∈ 0 j i → R := k [ τ¯ (e Ae )] B, ∩i∈Q0 i i ⊆ S := k [ τ¯ (e Ae )] B. ∪i∈Q0 i i ⊆ Theorem D. (Theorem 4.1.) Suppose τ : A ֒ M (B) is an impression of A with → |Q0| B an integral domain and τ(ei)= Eii for each i Q0. Then US/R = . Furthermore, if n Max S R is noetherian U , then∈ 6 ∅ { ∈ | n∩R } ⊆ S/R (1) The center Z of A is isomorphic to R and is depicted by S. (2) The statements (a) R = S. (b) A is a finitely generated Z-module. (c) Z is noetherian. (d) A is noetherian. satisfy the equivalences /' (a)sk 3+ (b)sk 3+ (c)og (d) ⋆ where (⋆) holds if the τ¯-image of a path is a monomial in B and I is generated by binomials in the paths of Q. Again consider the nonnoetherian algebra R = k + xS in example (1). In Example 4.3 and Proposition 4.4, we study the endomorphism ring A = End (R xS) R ⊕ of the reflexive R-module R xS. This algebra may be viewed as the noncommutative blowup of R at the isolated⊕ singular point xS of Max R [L, Section R]. Furthermore, NONNOETHERIAN GEOMETRY 5

A admits an impression τ : A ֒ M (S) satisfying → 2 R = k + xS = k [ τ¯ (e Ae )] and S = k[x, y]= k [ τ¯ (e Ae )] . ∩i∈Q0 i i ∪i∈Q0 i i By Theorem D, A is nonnoetherian, has center R, and is an inﬁnitely generated R- module. We show that the projective dimension and geometric codimension of its vertex simple modules Vi coincide:

pdA (Vi) = ght (anneiAei (Vi)) . This example suggests that the notion of geometric dimension is, in a suitable sense, a natural definition. Notation: We will denote by dim R the Krull dimension of R; by Frac R the ring of fractions of R; by Max R the set of maximal ideals of R; and by Spec R either the set of prime ideals of R or the affine k-scheme with global sections R. For a R we will denote by (a) either the closed set m Max R m a of Max R or⊂ the closed set p SpecZR p a of Spec R, depending{ ∈ on the| context.⊇ } For a subset Y of Max S,{ set∈Y c := Max| S⊇ Y}. \ In Section 4 we will denote by Q = (Q0, Q1, t, h) a quiver with vertex set Q0, arrow set Q , and head and tail maps h, t : Q Q . We will denote by kQ the 1 1 → 0 path algebra of Q, and by ei the idempotent corresponding to the vertex i Q0. Multiplication of paths is read right to left, following the composition of maps.∈ By module we mean left module. By infinitely generated R-module, we mean a module that is not finitely generated over R.

2. Nonnoetherian geometry as nonlocal geometry Throughout k is an algebraically closed field; S is an integral domain and a noetherian k-algebra; and R is a (possibly nonnoetherian) subalgebra of S. We begin with the following well known lemma. Lemma 2.1. If q Spec S, then q R Spec R. Furthermore, if S is a finitely generated k-algebra∈ and n Max S, then∩ n∈ R Max R. ∈ ∩ ∈ Proof. We show the second statement. Suppose S is a finitely generated k-algebra and let n Max S. Then S/n ∼= k since S is finitely generated over the algebraically .closed field∈ k.1 Thus the composition ψ : R ֒ S S/n is surjective since 1 R → → S ∈ Whence R/ ker ψ = k. Therefore ker ψ = n R is a maximal ideal of R. ∼ ∩ Consider the morphisms ι : Spec S Spec R and κ : Max S Spec R → → q q R n n R. 7→ ∩ 7→ ∩ 1This statement is false in general without the assumption that S is a finitely generated k-algebra. Indeed, let S be C, let k be the algebraic closure of Q, and let n be the maximal ideal 0 of C. Then S/n = k. 6∼ 6 CHARLIE BEIL

Let q Spec S and set p = ι(q)= q R. Then R p S q. Whence ∈ ∩ \ ⊆ \ (2) R = R S . ι(q) p ⊆ q Thus the embedding R ֒ S induces the morphism of schemes → ι : (Spec S, ) (Spec R, ) . OSpec S −→ OSpec R To aid our analysis of R, we introduce the following subsets of Max S and Spec S: U := n Max S R = S , S/R { ∈ | n∩R n} U˜S/R := q Spec S Rq∩R = Sq , (3) { ∈ | } U ∗ := n Max S R is noetherian , S/R { ∈ | n∩R } W := n Max S (n R)S = n . S/R ∈ | ∩ These subsets will play a centraln role throughoutp this paper.o Furthermore, US/R will play a central role in its sequels [B2, B3] in the context of dimer algebras. If R and S are ﬁxed, then we will often omit the subscript S/R. Lemma 2.2. Suppose p Spec R and q ι−1(p). If (q) U = , then R = S . ∈ ∈ Z ∩ 6 ∅ p q Proof. Let n (q) U and set m = n R. Then Rm = Sn. Furthermore, Rp Sq by (2). Thus,∈ since Z p∩ q, ∩ ⊆ ⊆ S =(S ) (S ) =(R ) = R S . q n q ⊆ n p m p p ⊆ q Therefore Rp = Sq. Lemma 2.3. Suppose q, q′ Spec S satisfy ∈ q R = q′ R and q q′. ∩ ∩ ⊆ If (q) U = or (q′) U = , then q = q′. Z ∩ 6 ∅ Z ∩ 6 ∅ ′ Proof. We claim that Sq = Sq′ , and consequently q = q since Sq has a unique maximal ideal. Indeed, suppose (q) U = . Then by Lemma 2.2, Z ∩ 6 ∅ S ′ S = R = R ′ S ′ . q ⊆ q q∩R q ∩R ⊆ q Therefore Sq = Sq′ . So suppose (q′) U = . Since q q′, we have Z ∩ 6 ∅ ⊆ (q′) (q). Z ⊆ Z Whence (q) U = , which was the previous case. Z ∩ 6 ∅ In the remainder of this section we assume that S is a ﬁnitely generated k-algebra, unless stated otherwise. Recall that S is an overring of a domain R if R S Frac R. ⊆ ⊆ NONNOETHERIAN GEOMETRY 7

Proposition 2.4. Suppose U is nonempty. Then (1) S is an overring of R; and (2) U is an open dense subset of Max S. In particular, the function fields of Spec S and Spec R coincide. Proof. (1) Suppose n U := U . Since S is an integral domain, we have ∈ S/R (4) Frac S = Frac(Sn) = Frac(Rn∩R) = Frac R. (2) We first claim that U coincides with the locus U ′ := n Max S S R . { ∈ | ⊆ n∩R} ′ Indeed, let n U. Then Sn = Rn∩R. Whence S Rn∩R. Thus U U . Conversely,∈ suppose n U ′. Then ⊆ ⊆ ∈ S = S(S n)−1 R (R (n R ))−1 = R = S . n \ ⊆ n∩R n∩R \ ∩ n∩R n∩R n ′ ′ Whence Rn∩R = Sn. Thus U U . Therefore U = U , proving our claim. Suppose that U = U ′ is nonempty.⊇ By assumption, S is a finitely generated k- algebra. Thus there is a finite set T := s1,...,sℓ S, minimal with respect to inclusion, such that { } ⊂

S = R [s1,...,sℓ] .

By Claim (1), there are elements r1,...,rℓ R such that risi R. Whence si −1 ∈ ∈ ∈ R ri . Therefore S R r−1,...,r−1 . ⊆ 1 ℓ Consider the open set ′′ UT := D(ri), 1≤\i≤ℓ where D(ri)= n Max S n ri is the principal open set where ri does not vanish. Then { ∈ | 6∋ } U ′′ U ′. T ⊆ Furthermore, U ′ is the union over all such minimal sets T S, ⊂ ′ ′′ U = UT . T [ In particular, U ′ is open. Thus U ′ is dense since S is an integral domain and U ′ is nonempty. Therefore U = U ′ is open dense. The morphism ι : Spec S Spec R is injective on the subset U˜ Spec S deﬁned in (3). Indeed, if q R = q′ →R and q, q′ U˜, then ⊆ ∩ ∩ ∈ Sq = Rq∩R = Rq′∩R = Sq′ . Whence q = q′. This fact is generalized in the following theorem. 8 CHARLIE BEIL

Furthermore, we will ﬁnd that the locus W Max S is similar in spirit to the Azumaya locus of A when A is a noncommutative⊂ algebra, module-ﬁnite over its center Z.2 Theorem 2.5. Let R be a subalgebra of S.

(1) If Rq∩R = Sq, then ι−1ι(q)= q . { } In particular, this holds if (q) US/R = . (2) The locus W is the subsetZ of∩ all n 6Max∅ S for which S/R ∈ κ−1κ(n)= n . { } In particular, U W . S/R ⊆ S/R (3) If US/R = , then Max S and Max R are isomorphic on open dense subsets, and thus6 birationally∅ equivalent. (4) If U = , then the Krull dimensions of R and S coincide, S/R 6 ∅ dim R = dim S. ′ −1 ′ Proof. (1.i) Let q U˜S/R, and suppose q ι ι(q). We want to show that q = q. Set p := q R =∈ q′ R. Then ∈ ∩ ∩ p =(q R) R q R . p ∩ p ⊆ q ∩ p ′ Thus, since pp is maximal in Rp, we have pp = qq Rp. Similarly pp = qq Rp. Whence ∩ ∩ (5) q R = p = q′ R . q ∩ p p q ∩ p In particular, since q U˜ , ∈ S/R US /R := n Max Sq (Rp) =(Sq) = qq . q p ∈ | n∩Rp n { } Furthermore, since q Un˜ and q′ ι−1ι(p), o ∈ S/R ∈ S = R S ′ . q p ⊆ q Thus q′ q. Therefore ⊆ (6) q′ q . q ⊆ q Whence (7) q′ U = q = . Z q ∩ Sq/Rp { q} 6 ∅ Thus, by Lemma 2.3, (5), (6), and (7) imply that ′ qq = qq. Therefore q′ = q′ S = q S = q, which is what we wanted to show. q ∩ q ∩ (1.ii) If (q) U = , then q U˜ by Lemma 2.2. Z ∩ S/R 6 ∅ ∈ S/R 2Recall that if n, n′ Max A and n Z = n′ Z is in the Azumaya locus of A, then n = n′. ∈ ∩ ∩ NONNOETHERIAN GEOMETRY 9

(2) We now claim that n W c if and only if there is a point n′ Max S, not ∈ S/R ∈ equal to n, such that κ(n)= κ(n′). First note that for any m Max R and n κ−1(m), we have m mS √mS n. ∈ ∈ ⊆ ⊆ ⊆ Thus m √mS R n R = m. Whence ⊆ ∩ ⊆ ∩ (8) √mS R = m. ∩ Now let n W c and set m := n R. Then by deﬁnition, n = √mS. Since S is ∈ S/R ∩ 6 Jacobson we have √mS = q. mS⊆q∈Max S \ Thus there exists a maximal ideal n′ = n of S such that √mS n′. Whence 6 ⊆ (8) κ(n)= n R = m = √mS R n′ R = κ(n′). ∩ ∩ ⊆ ∩ Therefore κ(n)= κ(n′) by Lemma 2.1. Conversely, suppose there are distinct points n, n′ Max S such that κ(n)= κ(n′). Then n R = n′ R =: m. Therefore ∈ ∩ ∩ √mS n n′ ( n. ⊆ ∩ (3) Suppose US/R = . By Claim (1) ι is injective on US/R. Furthermore, US/R is an open dense subset6 of∅ Max S by Proposition 2.4. Therefore Max S and Max R are birationally equivalent. (4) Finally, we show that dim R = dim S. Fix n U and set m := n R. Then ∈ S/R ∩ (i) (ii) (iii) (iv) dim R trdeg Frac R = trdeg Frac S = dim S = dim S = dim R dim R, ≤ k k n m ≤ where (i) holds by Proposition 2.4; (ii) and (iii) hold since S is a noetherian integral domain over k; and (iv) holds since n is in US/R. Example 2.6. Let S = k[x, y] and R = k[x, xy, xy2,...] = k + xS. For any b k, the ideals (x, y b)S,xS Spec S satisfy ∈ − ∈ (x, y b)S R = xS R =(x, xy, xy2,...) Max R. − ∩ ∩ ∈ Thus (x, y b)S W c by Theorem 2.5.2. − ∈ Remark 2.7. In general, U need not equal W . Indeed, consider the algebras S = k[x] and R = k + x2S = k[x2, x3] = k[u, v]/(u3 v2). ∼ − Then U = A1 0 and W = A1. \{ } The following proposition generalizes the fact that if R is a ﬁnitely generated k- algebra and m Max R, then ∈ R = k + m. Conversely, if I is an ideal in S and R = k + I, then (I) is a closed point in Spec R. In Corollary 3.14 below we will show that R = k + IZis nonnoetherian whenever the dimension of the subvariety (I) in Max S is nonzero. Z 10 CHARLIE BEIL

Proposition 2.8. Let S be an integral domain and a noetherian k-algebra. Consider a subalgebra R′ of S, an ideal I of S, and form the algebra R = k[R′,I]. Then U (hence W ) contains the open subset (I)c of Max S. Furthermore, if I S is a non-maximal radicalZ ideal and ⊂ R = k[I]= k + I, then U = W = (I)c. Z Proof. First suppose R = k[R′,I]. We claim that if q Spec S does not contain I, ∈ then Rq∩R = Sq; in particular, (9) (I)c U. Z ⊆ Set p := q R. Then R S . To show the reverse inclusion, suppose a S . ∩ p ⊆ q ∈ q Then there is some f,g S with g q such that a = f . Since q does not contain I, ∈ 6∈ g there is some c I q. Furthermore, since c,g S q and q is prime, we have ∈ \ ∈ \ cg S q. ∈ \ Since c I, we have cg I R. Whence ∈ ∈ ⊂ cg R p. ∈ \ But also b := agc = fc I R. ∈ ⊂ Thus a = b R . Therefore S R . Consequently R = S . cg ∈ p q ⊆ p q∩R q Now suppose R = k[I], where I is a non-maximal radical ideal of S. Let n (I). Then n I, so ∈ Z ⊇ n R I R = I. ∩ ⊇ ∩ Whence n R = I since I is a maximal ideal of R. But √IS = IS = I ( n since I is a radical∩ ideal of S. Thus (I) W c. Therefore Z ⊆ (i) (ii) (I) W c U c (I), Z ⊆ ⊆ ⊆ Z where (i) holds by Theorem 2.5.1 and (ii) holds by (9). Remark 2.9. U may properly contain (I)c; for example, take R′ = S, in which case U = Max S. Z Example 2.10. A geometric picture. (i) Let S = k[x, y] and R = k + xS. By Proposition 2.8, we can form Max R from 2 Max S = Ak by declaring the line (x)= x =0 A2 Z { } ⊂ k to be a single (closed) point, and all other points, U = x =0 , remain unaltered. In this description, (x) appears to be a 1-dimensional, hence{ 6 nonlocal,} point of Max R. Z NONNOETHERIAN GEOMETRY 11

(ii) Let S = k[x, y, z] and R = k[x,y,yz,yz2,...] = k[x,yS]. We can form Max R 3 from Max S = Ak by declaring each line (x c,y)= x = c,y =0 A3 Z − { } ⊂ k to be a single point, and all other points, U = y = 0 , remain unaltered. Similar to the previous example, each subvariety (x {c,y6 ) appears} to be a 1-dimensional, hence nonlocal, point of Max R. Z −

3. Depictions and the geometric dimension of a point Throughout S is an integral domain and a finitely generated k-algebra. We introduce the following definitions with the aim of constructing a geometric theory of nonnoetherian algebras, and in particular to formalize the geometric pic- tures in Example 2.10. Recall that if R is an integral domain and finitely generated over k, then the dimension of a point p Spec R coincides with the Krull dimension of R minus the height of p, ∈ dim p := dim R/p = dim R ht(p). − The dimension of p is then zero whenever p is a maximal ideal. Definition 3.1. A finitely generated k-algebra S is a depiction of a subalgebra R S if • ⊆ (i) the morphism ιS : Spec S Spec R, q q R, is surjective, (ii) for each n Max S, R →is noetherian7→ iff R∩ = S , and ∈ n∩R n∩R n (iii) US/R = . The geometric6 ∅ codimension or geometric height of p Spec R is the infimum • ∈ ght(p) := inf ht(q) q ι−1(p), S a depiction of R . | ∈ S The geometric dimension of p is the difference gdim p := dim R ght(p). − Remark 3.2. Note that condition (ii) is equivalent to U ∗ U . Furthermore, S/R ⊆ S/R this condition is independent of conditions (i) and (iii). Indeed, consider S = k[x, y], R = k[x, xy, y2,y3], and n =(x, y)S Max S. ∈ Then conditions (i) and (iii) hold.3 However, R is noetherian whereas R = S . n∩R n∩R 6 n Remark 3.3. Let S and S′ be depictions of R, and let p Spec R. Then in general −1 −1 ∈ the infimums of heights of ideals in ιS (p) and ιS′ (p) do not coincide. For example, let S = k[x, y, z], S′ = S[x−1], R = k + x(y, z)S, and m = x(y, z)S Max R. ∈ 3Naively it appears as though condition (i) may not hold since xy xS R xR, and so there is no q Spec S for which q R = xR. However, xR is not prime in R∈since∩x \y3 = (xy)y2. ∈ ∩ · 12 CHARLIE BEIL

Then inf ht(q) q ι−1 (m) = ht(xS)=1, | ∈ S whereas −1 ′ inf ht(q) q ι ′ (m) = ht((y, z)S )=2. | ∈ S Remark 3.4. In [S], Schwede gives a geometric description of subalgebras of noetherian algebras which is based on the gluing of schemes. In particular, the subalgebra 2 k[x, xy, xy ,...] k[x, y] is described as the ﬁber product k[x, y] k[y] k ([S, Example 3.7]). ⊂ × Lemma 3.5. If S and S′ are depictions of R, then

ιS (US/R)= ιS′ (US′/R). Proof. We claim that ∗ ∗ (10) ιS(US/R)= UR/R. Indeed, suppose m U ∗ . Then R is noetherian. Thus, since ι is surjective, ∈ R/R m S m ι (U ∗ ). Conversely, suppose m ι (U ∗ ). Then there is some n Max S for ∈ S S/R ∈ S S/R ∈ which n R = m and R is noetherian. Thus m U ∗ . Therefore (10) holds. ∩ n∩R ∈ R/R ′ ∗ ′ ∗ Since S and S are depictions of R, we have US/R = US/R and US /R = US′/R. Therefore by (10),

∗ ∗ ′ ∗ ′ ′ ιS(US/R)= ιS(US/R)= UR/R = ιS (US′/R)= ιS (US /R). The following lemma will be useful in Section 4. Lemma 3.6. Let R be a subalgebra of S, and suppose k is uncountable. Then the morphism ι : Spec S Spec R is surjective if and only if the morphism κ : Max S Max R is surjective. → → Proof. Suppose κ is surjective, and let p Spec R. Since S is a ﬁnitely generated k-algebra, R is a countably generated k-algebra.∈ By assumption k is uncountable, and thus R is Jacobson. Therefore, since p is prime, p = m. p⊆m∈Max R \ Since κ is surjective, the ideal q := n n∈ι−1(m) s.t. p⊆m\∈Max R satisﬁes q R = p. Furthermore, q is radical since it is the intersection of radical ideals. Thus,∩ since S is noetherian, the Lasker-Noether theorem implies q = q q , 1 ∩···∩ ℓ NONNOETHERIAN GEOMETRY 13 where each qi is a minimal prime over q. Set p := q R. Since p = q R q R = p , we have p p . Thus j j ∩ ∩ ⊆ j ∩ j ⊆ j p = pℓ √p p √p p = (q q ) R = q R = p. ⊆ 1 ··· ℓ ⊆ 1 ∩···∩ ℓ 1 ∩···∩ ℓ ∩ ∩ Whencep p p p = √p p . 1 ∩···∩ ℓ Therefore, since p is prime and each p is prime (Lemma 2.1), there is some 1 i ℓ j ≤ ≤ such that p = pi. Indeed, otherwise each pi would contain some ai not in p, but this is not possible since then a a p and p is prime.4 Therefore 1 ··· ℓ ∈ q R = p = p. i ∩ i It follows that ι is surjective. Lemma 3.7. Suppose S is a depiction of R. (1) If p Spec R and q ι−1(p), then ht (q) ht (p). (2) If m∈ Max R, then ht(∈ m) = dim R. ≤ ∈ Proof. (1) Let p Spec R, q ι−1(p), and set U := U . If (q) U = , then ∈ ∈ S/R Z ∩ 6 ∅ Sq = Rp by Lemma 2.2. Whence

ht (q) = dim Sq = dim Rp = ht(p). So suppose (q) U = . Since U = , U is an open dense set by Proposition 2.4. Therefore thereZ is a∩ maximal∅ chain of6 prime∅ ideals in S containing q, 0 ( q ( ( q ( q = q, 1 ··· ℓ−1 ℓ such that (qi) U = for each 1 i ℓ 1. Set p :=Zq R∩ Spec6 ∅ R. Then by≤ Lemma≤ − 2.2, R = S . In particular, i i ∩ ∈ pi qi ht (pℓ−1) = dim Rpℓ−1 = dim Sqℓ−1 = ht(qℓ−1) . Furthermore, by the contrapositive of Lemma 2.3,

pℓ−1 ( p. Therefore ht (q) ht (p). (2) Let m Max≤ R. Since ι is surjective, there is some q Spec S such that ι(q)= m. Let∈n be a maximal ideal of S containing q. Then m =∈q R n R, and so m = n R since m is maximal. Therefore ∩ ⊆ ∩ ∩ (i) (ii) (iii) dim R = dim S = ht(n) ht(m) dim R. ≤ ≤ 4 In general, p need not equal pj . Indeed, consider S = k[x, y], R = k + xS, and the prime ideals 2 p = (xy, xy ,...)R Spec R, q1 = xS Spec S, q2 = yS Spec S. ∈ ∈ ∈ Then p = (q1q2) R = xyS R = (xS R) (yS R)= (q1 R) (q2 R) ∩ ∩ ∩ ∩ ∩ ∩ and p2 := q2 R = p, but p1 := q1 R = p. ∩ ∩ 6 p p 14 CHARLIE BEIL

Indeed, (i) holds by Theorem 2.5.4; (ii) holds since S an integral domain; and (iii) holds by Claim (1). Theorem 3.8. Suppose R admits a depiction and let p Spec R. Then ∈ (11) ght(p) ht(p), ≤ −1 with equality if there is a depiction S of R and q ιS (p) such that (q) US/R = . Furthermore, ∈ Z ∩ 6 ∅ (1) If R is noetherian, then ght(p) = ht(p). (2) If m Max R, q ι−1(m), and dim (q) 1, then ∈ ∈ S Z ≥ ght(m) = ht(m). 6 Proof. The inequality (11) holds by Lemma 3.7.1. Furthermore, if (q) US/R = , then ght(p) = ht(p) by Lemma 3.5. Z ∩ 6 ∅ Suppose R is noetherian. Then R is a depiction of itself with UR/R = Max R. Therefore Claim (1) holds as a particular case of the previous paragraph. Now assume the hypotheses of Claim (2). Then

(i) (ii) (iii) (iv) ght(m) ht(q) < dim S = dim R = ht(m). ≤ Indeed, (i) holds by Definition 3.1; (ii) holds since dim (q) 1; (iii) holds by Theorem 2.5.4; and (iv) holds by Lemma 3.7.2. Z ≥ A depiction, by definition, is a finitely generated k-algebra. However, to capture a similar notion for local rings we introduce the following. Definition 3.9. Suppose S is a depiction of R and let q Spec S. Then we say S ∈ q is a local depiction of Rq∩R. Remark 3.10. A local depiction may not satisfy conditions (i) or (iii) in Definition 3.1. For example, consider S = k[x, y] and R = k + xS. Then the local ring SxS is a noetherian overring of the local ring RxS, but USxS /RxS = . Furthermore, although ι : Spec S Spec R is surjective,∅ the morphism ι : S → SxS Spec SxS Spec RxS is not surjective. Indeed, there are only two prime ideals of S contained→ in xS, (12) 0 ( xS. However, there are three prime ideals of R contained in xS, namely (13) 0 ( yS R ( xS R = xS. ∩ ∩ (These ideals are prime by Lemma 2.1.) Therefore ι : Spec S Spec R SxS xS → xS is not surjective. NONNOETHERIAN GEOMETRY 15

Example 3.11. We give an explicit example where the height and geometric codimension of an ideal do not coincide. Let S = k[x, y] and R = k + xS. Then the chains (12) and (13) imply ght(xS R)=1 =2=ht(xS R). ∩ 6 ∩ The following two theorems establish relationships between depictions, geometric dimension, and noetherianity. Theorem 3.12. Suppose S is a depiction of R. Then the following are equivalent: (1) R is noetherian. (2) US/R = Max S. (3) R = S. In particular, if R is noetherian, then its only depiction is itself. Proof. (2 1, 3) Suppose U = Max S. Then ⇒ S/R (i) (ii) (iii) (iv) R = Rm = Rn∩R = Sn = S. m n n ∈\Max R ∈Max\ S ∈Max\ S Indeed, (i) and (iv) hold since R and S are unital commutative rings; (ii) holds since S is a depiction of R, whence κ : Max S Max R is surjective; and (iii) holds since → US/R = Max S. Therefore R = S is noetherian. (2 3) Alternatively, suppose there is some g S R. Then the fractional ideal ⇒ ∈ \ (R : g)R is proper, and is thus contained in some maximal ideal m Max R. Let n κ−1(m). Then g S R . Thus n U c , and therefore U c =∈. ∈ ∈ n \ m ∈ S/R S/R 6 ∅ (1 2) Suppose R is noetherian. Then R is a depiction of itself. Let S be another depiction⇒ of R. Then (i) ι (U ) ι (Max S) Max R = ι (U ) = ι (U ), S S/R ⊆ S ⊆ R R/R S S/R where (i) holds by Lemma 3.5. Whence ιS (US/R) = ιS (Max S). But ιS is injective on US/R by Theorem 2.5.1. Therefore US/R = Max S. (3 1) If R = S, then R is trivially noetherian. ⇒ Theorem 3.13. If Max R contains a closed point of positive geometric dimension, then R is nonnoetherian. Conversely, if R is nonnoetherian, depicted by S, and there is some m ι(U c ) such that √mS = m, then Max R contains a closed point ∈ S/R of positive geometric dimension. Proof. (i) Suppose R contains a closed point of positive geometric dimension. Then R admits a depiction S. Let I be a radical ideal of S such that I R =: m is a maximal ideal of R and dim (I) 1. Let q Spec S be a minimal∩ prime over I. Then dim (q) 1. Furthermore,Z ≥m = I R ∈ q R = R implies m = q R since m is maximal.Z Thus≥ by Theorem 3.8.2, ∩ ⊆ ∩ 6 ∩ ght(m) = ht(m). 6 16 CHARLIE BEIL

Therefore R is not noetherian by Theorem 3.8.1. c (ii) Suppose the hypotheses hold, and dim US/R = 0. We claim that R is noetherian. More specifically, we claim that R is a finitely generated k-algebra. To show this, it suffices to show that S is a finitely generated R-module by the Artin-Tate lemma. By the Lasker-Noether theorem, there are ideals n ,..., n Spec S such that 1 ℓ ∈ m = √mS = n n . 1 ∩···∩ ℓ Whence m = m R =(n n ) R =(n R) (n R). ∩ 1 ∩···∩ ℓ ∩ 1 ∩ ∩···∩ ℓ ∩ Since m is maximal in R and each n R is proper, we have i ∩ n R = = n R = m. 1 ∩ ··· ℓ ∩ c Therefore, since dim US/R = 0, each ni is a maximal ideal of S. Let x1,...,xt be a minimal generating set for S over k. Since k is algebraically closed, for each 1 i ℓ and 1 j t there are scalars α k such that ≤ ≤ ≤ ≤ ij ∈ (14) x α n . j − ij ∈ i Set the degree of each x to be 1. Denote by the set of monomials in the variables j M x1,...,xt, with coefficient 1 and degree at most ℓ 1. Since S is a finitely generated k-algebra, is finite. − We proceedM by induction to show that as k-spaces, (15) S = R + km. mX∈M Base case. Consider a monomial x x of degree ℓ. Set j1 ··· jℓ r := (x α )(x α ) (x α ) and h := x x r. j1 − 1j1 j2 − 2j2 ··· jℓ − ℓjℓ j1 ··· jℓ − Then r is in n1 nℓ = m R by (14). Furthermore, h has degree at most ℓ 1. Thus h is in ∩···∩km by the⊂ definition of . Therefore x x = r + h is− in m∈M M j1 ··· jℓ R + m∈M km. Inductive step.P Now suppose all monomials of degree at most d 1 are in R + P km, and consider a monomial x x of degree d. Set − m∈M j1 ··· jd r := (x α ) (x α ) x x and h := x x r. P j1 − 1j1 ··· jℓ − ℓjℓ jℓ+1 ··· jd j1 ··· jd − Then r is in n1 nℓ = m R, again by (14). Furthermore, h has degree at most d 1. Thus∩···∩h is in R + ⊂ km by the induction hypothesis. Therefore − m∈M xj1 xjd = r + h is in R + m∈M km. This proves our claim (15). But··· P RP+ km R + Rm. ⊆ m∈M m∈M X X Thus as R-modules, S = R + Rm. mX∈M NONNOETHERIAN GEOMETRY 17

Therefore S is a finitely generated R-module since is finite. M The following corollary is immediate. Corollary 3.14. Let I be a radical ideal of a finitely generated k-algebra S. Then the ring R = k + I is nonnoetherian if and only if dim (I) 1. Z ≥ Example 3.15. Let S be a finitely generated k-algebra, and let n1, n2,..., nℓ be a finite set of maximal ideals of S. Then by Corollary 3.14, the ring R = k + √n n n 1 2 ··· ℓ is noetherian. Furthermore, the ℓ points n1,..., nℓ of Max S are identified as one point in Max R, and all other points are unaltered by Proposition 2.8, U = (n n )c. S/R Z 1 ··· ℓ In particular, the variety Max R is nonlocal. R is therefore an example of a noetherian ring with nonlocal geometry, although its one nonlocal point is zero-dimensional. The following proposition characterizes the assumptions in the converse implication of Theorem 3.13. Proposition 3.16. If there is a point m ι(U c ) satisfying mS = m, then ∈ S/R (16) U c = m . S/R { } If additionally S is a depiction of R, then R is an isolated nonnoetherian singularity. Proof. Suppose m = mS. Consider m′ Max R m . Fix n′ ι−1(m′) and g m m′. Since m = mS, the ideal Sg is in m ∈R. Thus\{ } ∈ ∈ \ ⊂ −1 S = Sg g R ′ . · ⊂ m ′ Whence Rm′ = Sn′ . Thus m ι(US/R). Therefore (16) holds. ∈ ∗ If S is a depiction of R, then US/R = US/R, and so (16) implies that R is an isolated nonnoetherian singularity. Question 3.17. Is there a nonnoetherian algebra (which admits a depiction) such that all of its closed points have geometric dimension zero? By Proposition 3.16, such an algebra would necessarily be a non-isolated nonnoetherian singularity. Or is it the case that R is nonnoetherian if and only if Max R contains a closed point of positive geometric dimension? Example 3.18. Consider the algebras S = k [x, y] , R = k + xS, R′ = k [x, xy] . Set n := (x, y)S. By Example 3.11, the closed point n R of Max R has geometric dimension 1. Naively it appears that the closed point n R∩′ of Max R′ should also have geometric dimension 1, contrary to the claim of Theorem∩ 3.13 since R′ is noetherian. 18 CHARLIE BEIL

However, although R is depicted by S, R′ is not. Indeed, the morphism ι : Spec S Spec R′ is not surjective: the ideal xR′ is prime in R′, but → ι−1(xR′)= . ∅ In contrast, the ideal xR is not prime in R since x xy2 = xy xy. In view of Lemma 3.6, consider the maximal ideals· · m := (x, xy α)R′ Max R′ α − ∈ with α k. Then ∈ xR′ = m . ∩α∈k α Furthermore, if α = 0, then 1 m S. Thus if n Max S satisﬁes n R = m , then 6 ∈ α ∈ ∩ α 1 m S n, ∈ α ⊆ which is not possible. Therefore the morphism κ : Max S Max R′ is also not surjective. → We say a depiction S of R is maximal (resp. minimal) if S is not contained in (resp. does not contain) any other depiction of R. Proposition 3.19. (1) Minimal depictions do not exist in general. (2) Maximal depictions are not unique in general. Proof. (1) We ﬁrst show that minimal depictions need not exist. Let S = k[x, y] and R = k + xS. For ℓ N, set ∈ ℓ ℓ+1 ℓ+2 2 ℓ−1 ℓ ℓ+1 2ℓ−1 Sℓ := R y ,y ,y ,... = k x, xy, xy ,...,xy ,y ,y ,...,y .

Then each Sℓ is a depiction of R . However,

Sℓ+1 ( Sℓ and R = Sℓ. ℓ\≥1 (2) We now show that maximal depictions need not be unique. Let T = k[x, y, z] and R = k + xyT . We claim that the overrings S := T [x−1] and S′ := T [y−1] are both depictions of R. Indeed, U is nonempty: Let a, b k∗. Then x−1 and each f T are in S/R ∈ ∈ R(x−a,y−b,z)S∩R since 1 1 x−1 = xy2 and f = xyf . · (xy)2 · xy Thus R(x−a,y−b,z)S∩R = S(x−a,y−b,z)S. Therefore the maximal ideal (x a, y b, z)S is in U . Similarly U ′ is nonempty. − − S/R S /R NONNOETHERIAN GEOMETRY 19

′ It is straightforward to check that ιS : Spec S Spec R and ιS′ : Spec S Spec R are both surjective by Lemma 3.6 and noting that→ → yS R = xS′ R = xyT Max R. ∩ ∩ ∈ Finally, the minimal proper overring T [x−1,y−1] of S and S′ is not a depiction of R: T [x−1,y−1] has no maximal ideal n satisfying n R = xyT . Therefore ∩ −1 −1 ι −1 −1 : Spec T x ,y Spec R T [x ,y ] → is not surjective.

4. Noncommutative nonnoetherian geometry Throughout k is an uncountable algebraically closed field of characteristic zero; A = kQ/I is a quiver algebra with finite quiver Q and I kQ≥1; and B is a finitely generated integral domain over k. We say an element c ⊂A = kQ/I is a cycle (resp. path) if there is a cycle (resp. path) c′ kQ such that c∈= c′ + I. ∈ Denote by Eji M|Q0|(B) the matrix with a 1 in the (ji)-th slot and zeros elsewhere. Let τ : A∈ M (B) be an algebra homomorphism such that τ(e ) = E → |Q0| i ii for each i Q0. For p ejAei, denote byτ ¯(p) B the single non-zero entry of the matrix τ(p∈). For each i,∈ j Q ,τ ¯ defines a k-linear∈ mapτ ¯ : e Ae B. Set ∈ 0 j i → R := k [ τ¯ (e Ae )] B, ∩i∈Q0 i i ⊆ S := k [ τ¯ (e Ae )] B. ∪i∈Q0 i i ⊆ Recall the definition of an impression given in Definition 1.1. An impression is useful in part because it determines the center Z of A [B, Lemma 2.1]: (17) Z = f B f1 im τ B. ∼ { ∈ | d ∈ } ⊆

Theorem 4.1. Suppose τ : A ֒ M|Q0|(B) is an impression of A with B an integral domain and τ(e ) = E for each→ i Q . Then U = . Furthermore, if U ∗ i ii ∈ 0 S/R 6 ∅ S/R ⊆ US/R, then (1) The center Z of A is isomorphic to R and is depicted by S. (2) The statements (a) R = S. (b) A is a ﬁnitely generated Z-module. (c) Z is noetherian. (d) A is noetherian. satisfy the equivalences /' (a)sk 3+ (b)sk 3+ (c)og (d) ⋆ where (⋆) holds if the τ¯-image of a path is a monomial in B and I is generated by binomials in the paths of Q. 20 CHARLIE BEIL

In Example 4.5 below we show that the assumption U ∗ U is independent S/R ⊆ S/R from the impression assumption.

Proof. (1.i) We first show that the center of A is isomorphic to R. Set d := Q0 . Clearly | | f B f1 im τ R. { ∈ | d ∈ } ⊆ So suppose f R. Then for each i Q there is some c e Ae such thatτ ¯(c )= f. ∈ ∈ 0 i ∈ i i i Thus τ c = f1 . Whence f B f1 im τ R. Thus by (17), i∈Q0 i d { ∈ | d ∈ } ⊇ P Z = f B f1 im τ = R. ∼ { ∈ | d ∈ } (1.ii) We now show that R is depicted by S. (1.ii.i) S is a finitely generated domain over k. By assumption Q is finite. Thus there are a finite number of non-vertex cycles in Q without cyclic proper subpaths (not viewed modulo I), say c1,...,cℓ. Note that each cycle has length at most Q0 . We claim that | |

(18) S = k [¯τ (c1) ,..., τ¯ (cℓ)] . It suﬃces to show that theτ ¯-image of each non-vertex cycle with a cyclic proper subpath is contained in k [¯τ (c1) ,..., τ¯ (cℓ)] . So suppose d is such a cycle. Then d has a cyclic proper subpath c with no cyclic proper subpaths (again, not viewed modulo I). Thus there are paths d1, d2 such that d = d2cd1. Since c is a cycle, d2d1 is also a cycle. Therefore, since τ is an algebra homomorphism and B is commutative,

τ¯(d)=¯τ (d2cd1)=¯τ (d2)¯τ (c)¯τ (d1)=¯τ (c)¯τ (d2)¯τ (d1)=¯τ (c)¯τ (d2d1) .

The length of the cycle d2d1 is strictly less than the length of d since c is a non-vertex cycle. Our claim (18) then follows by induction on the length of the cycles. Furthermore, S is a domain since it is a subalgebra of the domain B. (1.ii.ii) The set US/R is nonempty. By [B, Lemma 2.4], the dimension vector for the Q0 simple A-modules of maximal k-dimension is 1 . Thus there exists a path pji I between any two vertices i, j of Q. The cycle b := p p p then contains each6∈ 1d ··· 32 21 vertex as a subpath. Furthermore, since τ is injective,τ ¯(pji) = 0. Thus, since B is an integral domain and τ is an algebra homomorphism, 6 τ¯(b)=¯τ (p p p )=¯τ (p ) τ¯ (p )¯τ (p ) =0. 1d ··· 32 21 1d ··· 32 21 6 Whenceτ ¯(b) = 0. Fix i Q ,6 and let c e Ae be an arbitrary cycle. For each j Q , denote by b ∈ 0 i ∈ i i ∈ 0 j and dj the respective cycles obtained by cyclically permuting b and di := bci so that their heads and tails are at j. Thenτ ¯(bj)=¯τ(b) =: β andτ ¯(dj)=¯τ(di)=¯τ(ci)β, since τ is an algebra homomorphism on eiAei. Therefore β andτ ¯(ci)β are in R. Fix b (β)c Max B. By Lemma 2.1, n := b S and m := n R are maximal ideals of∈S Zand R⊂respectively. Furthermore, β R∩ is invertible in∩ the localization ∈ NONNOETHERIAN GEOMETRY 21

Rm. Consequently τ¯(c )=¯τ(c )β β−1 R . i i · ∈ m Since ci was arbitrary, Sn Rm. Whence Sn = Rm. (1.ii.iii) The map ι : Spec⊆ S Spec R, q q R, is surjective. By Claim (1.ii.a), S is a finitely generated k→-algebra. Thus7→ by∩ Lemma 3.6, it suffices to show κ : Max S Max R, n n R, is surjective. Let m Max R. By the definition of → 7→ ∩ ∈ impression, the morphism Max B Max τ(Z), b b1d τ(Z), is surjective. Thus there is some b Max B such that→b R = m. By Lemma7→ ∩ 2.1, n := b S is in Max S. Furthermore, n∈satisfies ∩ ∩ n R =(b S) R = m. ∩ ∩ ∩ Thus κ is surjective. (1.ii.iv) Claims (1.ii.i, ii, iii), together with the assumption U ∗ U , imply S/R ⊆ S/R that S is a depiction of R. (2. a c) Follows from Claim (1) and Theorem 3.12.2. (2. a ⇔ b,d) Follows from [B, Theorem 2.11]. (2. b ⇒ a) Suppose R = S. Then R is an infinitely generated k-algebra by Theorem 3.12.1.⇒ Furthermore, S 6 is a finitely generated k-algebra by Claim (1.ii.i). Thus S is an infinitely generated R-module by the Artin-Tate lemma. But τ is injective, Q < , and Z = R by Claim (1). Therefore e Ae is an infinitely generated 0 ∼ i∈Q0 i i |Z-module.| ∞ Whence A is an infinitely generated Z-module. (2. d a) Suppose R = S, and the conditionsL (⋆) hold. As was shown in (b a), S is an⇒ infinitely generated6 R-module. Thus there is a cycle q and vertex i such⇒ that for each n 1, ≥ τ¯(q)n S τ¯ (e Ae ) . ∈ \ i i Since τ is an impression of A, there is a path p1 in et(q)Aei and a path p2 in eiAet(q). Assume to the contrary that the chain of ideals 0 ( (p p ) ( (p p ,p qp ) ( p p ,p qp ,p q2p ( 2 1 2 1 2 1 2 1 2 1 2 1 ··· in eiAei terminates. Then, since I is generated by binomials in the paths of Q, there is some n > m 0 and a cycle a e Ae such that ≥ ∈ i i n m p2q p1 = ap2q p1. Whence n n m m τ¯(q) τ¯ (p2p1)=¯τ (p2q p1)=¯τ (ap2q p1)=¯τ(a)¯τ(q) τ¯ (p2p1) . Thus since B is an integral domain, τ¯(q)n−m =τ ¯(a) τ¯ (e Ae ) . ∈ i i But this contradicts our choice of q. Thus the vertex corner ring eiAei is nonnoetherian. Therefore A is nonnoetherian. 22 CHARLIE BEIL

a y 2) 1 9 i b

Figure 1.

Remark 4.2. The role of S is new: S is a commutative ring obtained from A that in most cases is not a central subring of A, but is closely related to the geometry of the center Z of A. By Theorem 4.1.2, if Z is noetherian, then S is isomorphic to Z, and if Z is nonnoetherian, then S properly contains Z. Example 4.3. Consider the quiver algebra (19) A = kQ/ yba bay h − i with quiver given in Figure 1. A admits the impression (τ, B = k[x, y]) where τ(ei)= Eii for i =1, 2, and (20)τ ¯(a)=1, τ¯(b)= x, τ¯(y)= y. Recall our motivating example in Section 2, S′ = k[x, y] and R′ = k + xS′. By (17), the center of A is isomorphic to R = k [¯τ (e Ae ) τ¯ (e Ae )] = k + xS′ = R′ 1 1 ∩ 2 2 and is depicted by S = k [¯τ (e Ae ) τ¯ (e Ae )] = k[x, y]= S′. 1 1 ∪ 2 2 By Theorem 4.1.2, A and Z are nonnoetherian, and A is an inﬁnitely generated Z-module. We recall a homological characterization of smoothness, in the both the commutative and noncommutative settings. Let R be a noetherian integral domain and let p Spec R. Then by the Auslander-Buchsbaum formula, ∈ ht(p) if p is smooth pdRp (Rp/p)= ( otherwise. ∞ This notion was generalized to the noncommutative setting by Brown and Hajarnavis [BH]. They deﬁne a noetherian (noncommutative) algebra A with prime center R to be homologically homogeneous if the projective dimension of each simple A-module V equals the Krull dimension of R,5

(21) pdA(V ) = dim R = ht(annR(V )).

5Speciﬁcally, if R is a commutative noetherian equidimensional k-algebra and A is a module-ﬁnite R-algebra, then A is homologically homogeneous if all simple A-modules have the same projective dimension. NONNOETHERIAN GEOMETRY 23

Using this notion, Van den Bergh defines a noncommutative crepant resolution A of a noetherian normal Gorenstein domain R to be a homologically homogeneous endomorphism ring A = EndR(M) of a reflexive finitely generated R-module M [V, Definition 4.1]. We propose that if R is a nonnoetherian domain, then homological homogeneity should be replaced by an equality between projective dimension and geometric height, rather than height as in (21). This proposal is illustrated in the following proposition, and is further studied in [B4] in the context of homotopy dimer algebras. Proposition 4.4. Let A = kQ/I be the quiver algebra (19), and set m = xS. Then A is an endomorphism of a reflexive module over its center Z = R, • ∼ A = End (R m) = End (Ae ) . ∼ R ⊕ ∼ Z 1 Let V be the simple A-module supported at vertex i. Then • i pdA(Vi) = ght (anneiAei (Vi)) . The smooth locus of Z parameterizes the simple A-module isoclasses of max- • imal k-dimension, and coincides with the open set U Max S. S/R ⊂ Therefore, although A and Z are nonnoetherian and A is an infinitely generated Z- module, A nevertheless may be viewed as a noncommutative desingularization of its center.

Proof. For the following, denote HomR( , ) by( , ). Using the labeling of arrows given by the impression (20), we ﬁnd − − − − End (Ae ) = End (e Ae e Ae ) Z 1 Z 1 1 ⊕ 2 1 = End (R m) ∼ R ⊕ (R, R) (m, R) ∼= (R, m) (m, m)! R S ∼= m S!

e1Ae1 e1Ae2 ∼= e2Ae1 e2Ae2! ∼= A. Furthermore, R m is a reﬂexive R-module: ⊕ ((R m, R), R) = (R S, R) = R m. ⊕ ∼ ⊕ ∼ ⊕ The minimal projective resolution of V1 is

S ·x R R/m k 0 ∼= = V1 0. → S! −→ m! → 0 ! 0! → 24 CHARLIE BEIL

Set n := (x, y)S Max S. Then the minimal projective resolution of V is ∈ 2

1 · ·   S xy x R S y S 0 0 −   0 ∼= = V2 0. → S! −→ m S! −→ S! → S/n! k! → The simple A-modules of maximal k-dimension are the simples modules with dimension vector (1, 1) by [B, Lemma 5.1]. These modules are parameterized by the ∗ smooth locus of Max Z, namely (ab, y) k k, which coincides with US/R Max S. ∈ × ⊂

Although A is isomorphic to EndZ (Ae1), note that A is not isomorphic to EndZ (Ae2) since End (Ae ) = End (S S) = M (S). Z 2 ∼ R ⊕ ∼ 2 Furthermore, the moduli space of θ-stable A-modules of dimension vector (1, 1), for generic stability parameter θ, is precisely the desingularization Max S. (Max S is not a resolution of Max R since the morphism κ : Max S Max R is not proper.) The following example demonstrates the necessity→ of the assumption U ∗ U S/R ⊆ S/R in Theorem 4.1. Example 4.5. Consider the quiver algebra A = kQ/ yba bay, y2 ba − − with quiver given in Figure 1, as in Example 4.3. A admits an impression (τ, B = k[x]) where τ(ei)= Eii for i =1, 2, and τ¯(a)=¯τ(b)=¯τ(y)= x. By (17), the center Z of A is isomorphic to R = k x2, x3 . Therefore Z is noetherian. But S = k [x] = R. 6 However, Theorem 4.1.2 is not applicable to this example because U ∗ = A1 A1 0 = U ; S/R 6⊆ \{ } S/R see Remark 2.7.

Acknowledgments. The author would like to thank Paul Smith and Travis Kopp for useful discussions, and an anonymous referee for useful comments. Part of this paper is based on work supported by the Simons Foundation while the author was a postdoc at the Simons Center for Geometry and Physics at Stony Brook University. His work was also supported in part by a PFGW Grant, which he gratefully acknowledges. NONNOETHERIAN GEOMETRY 25

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Heilbronn Institute for Mathematical Research, School of Mathematics, Howard House, The University of Bristol, Bristol, BS8 1SN, United Kingdom. E-mail address: [email protected]